You are designing a hydroelectric plant. The turbine has an efficiency of 88% and the generator...
You are designing a hydroelectric plant. The turbine has an efficiency of 88% and the generator has an efficiency of 92%. You design the water flow rate to be 110 m3/s. The water elevation drop is 120 m. Disregarding frictional loss in the piping, what do you expect the electric power output to be in MW?
Homework Answers
Answer: Given, volumetric flow rate of water = 110 m3/s
So, mass flow rate = volumetric flow rate*density = 110 m3/s*1000 Kg/m3 = 1.1*105 Kg/s
Elevation drop = 120 m
So, actual electrical power output = mass flow rate* acceleration due to gravity*Elevation drop* generator efficiency*turbine efficiency
So, P = (1.1*105 Kg/s)*(9.8 m/s2)*(120 m)*0.92*0.88 = 104.73*106 W = 104.73 MW.
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